Kinematics

JEE Main level, single correct, medium

44 of 308

Question 44 of 308

A body starts moving from rest with constant acceleration covers displacement S1{S}_{1} in first (p−1)(p-1) seconds and S2{S}_{2} in first pp seconds. The displacement S1+S2{S}_{1}+{S}_{2} will be made in time :

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Answer: (B) (2p2−2p+1)s\sqrt{\left(2{p}^{2}-2p+1\right)}s

For the displacement S1{S}_{1} of the body, it can be written that
S1=12a(p−1)2...(1){S}_{1}=\dfrac{1}{2}a(p-1{)}^{2}...\left(1\right)
And, for the displacement S2{S}_{2} of the body, it follows that
S2=12a(p)2...(2){S}_{2}=\dfrac{1}{2}a(p{)}^{2}...\left(2\right)
If tt be the required time, it follows that
S1+S2=12at2...(3){S}_{1}+{S}_{2}=\dfrac{1}{2}a{t}^{2}...\left(3\right)
Equations (1), (2) and (3) imply that
(p−1)2+p2=t2⇒t=2p2−2p+1s(p-1{)}^{2}+{p}^{2}={t}^{2} \Rightarrow t=\sqrt{2{p}^{2}-2p+1}s

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